170 ANALYTICAL MECHANICS
where dG is the moment of dF about the axis. Thus the total work done by all the forces acting upon all the elements in producing the angular displacement dd is
°dG de
-r°
Jo
= dd f Jo
? (de is the same for every element of mass.)
where G is the sum of the moments about the axis of all the forces acting upon the elements of the body, i.e., the resultant torque about the axis. The work done in giving the body a finite angular displacement is, therefore,
W= Gde (III)
Jo
= GO [when G is constant],
Therefore work done by a constant torque in producing an angular displacement equals the product of the torque by the angle.
PROBLEMS.
1. A weight of 10 tons is to be raised by a jackscrew. The pitch of the screw is $ inch and the length of the bar which is used to turn the nut on the screw is 2 feet long. Supposing the work done by the torque to be expended entirely against gravitational forces, find the force which must be applied at the end of the bar.
2. A ball, which is suspended by a string of negligible mass, is pulled aside until the string makes an angle 6 with a vertical line. Show that the work done is the same whether it is supposed to have been done in raising the ball against the action of gravitational forces, or in rotating the ball and the string, as a whole, about a horizontal axis through the point of suspension, against the action of the torque.
3. In the preceding problem take the following data and calculate, by both methods, the amount of work done. Weight of ball = 12 ounces, length of string = 3 feet, and 0 = 60°.
4. The torque which has to be applied to the ends of a rod varies directly with the angle through which it is twisted; derive an expression for the work done in turning one end of the rod with respect to the other end through an angle 6.